Bird R.B., Stewart W.E., Lightfoot E.N. Transport Phenomena
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510.9 Free Convection
317
fluid in the upward-moving stream is the same as that in the downward-moving
stream. The plates are presumed to be very tall, so that end effects near the top and bot-
tom can be disregarded. Then for all practical purposes the temperature is a function of
y
alone.
An energy balance can now be made over a thin slab of fluid of thickness
Ay,
using
the y-component of the combined energy flux vector
e
as given in Eq. 9.8-6. The term
containing the kinetic energy and enthalpy can be disregarded, since the y-component of
the
v
vector is zero. The y-component of the term
[T
.
vl
is
rYv,
=
-p(dv,/dy)v,, which
would lead to the viscous heating contribution discussed in 510.4. However, in the very
slow flows encountered in free convection, this term will be extremely small and can be
neglected. The energy balance then leads to the equation
for constant
k.
The temperature equation is to be solved with the boundary conditions:
B.C.
1:
B.C.
2:
The solution to this problem is
in which AT
=
T,
-
TI is the difference of the wall temperatures, and
=
+(T,
+
T2) is
their arithmetic mean.
By making a momentum balance over the same slab of thickness Ay, one arrives at a
differential equation for the velocity distribution
Here the viscosity has been assumed constant (see Problem 10B.ll for a solution with
temperature-dependent viscosity).
The phenomenon of free convection results from the fact that when the fluid is
heated, the density (usually) decreases and the fluid rises. The mathematical description
of the system must take this essential feature of the phenomenon into account. Because
the temperature difference AT
=
T2
-
TI is taken to be small in this problem, it can be ex-
pected that the density changes in the system will be small. This suggests that we should
expand
p
in a Taylor series about the temperature
T
=
f
(T,
+
T2) thus:
Here
5
and
p
are the density and coefficient of volume expansion evaluated at the tem-
perature
T.
The coefficient of volume expansion is defined as
We now introduce the "Taylor-made" equation of state of Eq. 10.9-6 (keeping two terms
only) into the equation of motion in Eq. 10.9-5 to get
318
Chapter
10
Shell Energy Balances and Temperature Distributions
in
Solids and Laminar Flow
This equation describes the balance amonx the viscous force, the pressure force, the
gravity force, and the buoyant force
-&3(T
-
T)
(all per unit volume). Into this we
now substitute the temperature distribution given in Eq.
10.9-4
to get the differential
equation
which is to be solved with the boundary conditions
B.C.
1:
B.C.
2:
aty=
-B,
v,=O
aty
=
+B,
v,
=
0
The solution is
We now require that the net mass flow in the
z
direction be zero, that is,
Substitution of
v,
from Eq.
10.9-12
and
p
from Eqs.
10.9-6
and
4
into this integral leads
to
the conclusion that
when terms containing the square of the small quantity
AT
are neglected. Equation
10.9-
14
states that the pressure gradient
in
the system is due solely to the weight of the fluid,
and the usual hydrostatic pressure distribution prevails. Therefore the second term
on
the right side of Eq.
10.9-12
drops out and the final expression for the velocity distribu-
tion is
The average velocity in the upward-moving stream is
The motion of the fluid is thus a direct result of the buoyant force term in Eq.
10.9-8,
as-
sociated with the temperature gradient in the system. The velocity distribution of
Eq.
10.9-15
is shown in Fig.
10.9-1.
It is this sort of velocity distribution that occurs in the
air
space in a double-pane window or in double-wall panels in buildings. It is also this
kind
of flow that occurs in the operation of a Clusius-Dickel column used for separating iso-
topes or organic liquid mixtures by the combined effects of thermal diffusion and free
convection.'
Thermal diffusion is the diffusion resulting from
a
temperature gradient. For
a
lucid discussion
of the Clusius-Dickel column see
K.
E.
Grew
and
T.
L.
Ibbs,
Thermal
Diffusion
in
Gases,
Cambridge
University Press
(1952),
pp.
94-106.
Questions for Discussion
319
The velocity distribution in
Eq.
10.9-15 may be rewritten using a dimensionless ve-
locity
6,
=
Bv,p/p
and a dimensionless coordinate
5
=
y/B
thus:
Here Gr is the dimensionless
Gmshof
number:
defined by
where
Ap
=
p,
-
p,.
The second form of the Grashof number is obtained from the first
form by using
Eq.
10.9-6. The Grashof number is the characteristic group occurring in
analyses of free convection, as is shown by dimensional analysis in Chapter
11.
It arises
in heat transfer coefficient correlations in Chapter
14.
QUESTIONS FOR DISCUSSION
1.
Verify that the Brinkman, Biot, Prandtl, and Grashof numbers are dimensionless.
2.
To what problem in electrical circuits is the addition of thermal resistances analogous?
3.
What is the coefficient of volume expansion for an ideal gas? What is the corresponding ex-
pression for the Grashof number?
4.
What might be some consequences of large temperature gradients produced by viscous heat-
ing in viscometry, lubrication, and plastics extrusion?
5.
In 510.8 would there be any advantage to choosing the dimensionless temperature and di-
mensionless axial coordinate to be
O
=
(T
-
T1)/Tl
and
{
=
z/R?
6.
What would happen in
s9.9
if
the fluid were water and
T
were 4"C?
7.
Is there any advantage to solving
Eq.
9.7-9 in terms of hyperbolic functions rather than expo-
nential functions?
8.
In going from Eq. 10.8-11 to Eq. 10.8-12 the axial conduction term was neglected with respect
to the axial convection term. To justify this, put in some reasonable numerical values to esti-
mate the relative sizes of the terms.
9.
How serious is it to neglect the dependence of viscosity on temperature in solving forced con-
vection problems? Viscous dissipation heating problems?
10.
At steady state the temperature profiles in a laminated system appear thus:
Which material has the higher thermal conductivity?
Distance
-
11.
Show that
Eq.
10.6-4 can be obtained directly by rewriting Eq. 10.6-1 with
x
+
Ax
replaced by
x,.
Similarly, one gets Eq. 10.6-20 from Eq. 10.6-17, with
r
+
Ar
replaced by
r,.
Named for Franz Grashof (1826-1893) (pronounced "Grahss-hoff). He was professor of applied
mechanics in Karlsruhe and one of the founders of the Verein Deutscher Ingenieure in 1856.
320
Chapter
10
Shell Energy Balances and Temperature Distributions in Solids and Laminar Flow
PROBLEMS
10A.1.
Heat loss from an insulated pipe. A standard schedule
40,2-in.
steel pipe (inside diameter
2.067
in. and wall thickness
0.154
in.) carrying steam is insulated with
2
in. of
85%
magnesia
covered in turn with
2
in. of cork. Estimate the heat loss per hour per foot of pipe if the inner
surface of the pipe is at
250°F
and the outer surface of the cork is at
90°F.
The thermal
conductivities (in Btu/hr
ft
.
F)
of the substances concerned are: steel,
26.1; 85%
magnesia,
0.04;
cork,
0.03.
Answer:
24
Btu/ hr
-
ft
10.A.2.
Heat loss from a rectangular fin. Calculate the heat loss from a rectangular fin (see Fig.
10.7-1)
for the following conditions:
Air temperature
Wall temperature
Thermal conductivity of fin
Thermal conductivity of air
Heat transfer coefficient
Length of fin
Width of fin
Thickness of fin
Answer:
2074
Btu/hr
350°F
500°F
60
Btu/hr. ft
.F
0.0022
Btu/hr ft F
120
Btu/hr
ft?
.
F
0.2
ft
1.0
ft
0.16
in.
10A.3.
Maximum temperature in a lubricant. An oil is acting as a lubricant for a pair of cylindrical
surfaces such as those shown in Fig.
10.4-1.
The angular velocity of the outer cylinder is
7908
rpm. The outer cylinder has a radius of
5.06
cm, and the clearance between the cylinders
is
0.027
cm. What is the maximum temperature in the oil if both wall temperatures are known
to
be
15S°F?
The physical properties of the oil are assumed constant at the following values:
Viscosity
92.3
cp
Density
1.22
g/cm3
Thermal conductivity
0.0055
cal/s
.
cm
-
C
Answer:
174°F
10A.4.
Current-carrying capacity of wire.
A
copper wire of
0.040
in. diameter is insulated uni-
formly with plastic to an outer diameter of
0.12
in. and is exposed to surroundings at
100°F.
The heat transfer coefficient from the outer surface of the plastic to the surroundings is
1.5
Btu/hr
ft?
.
F.
What is the maximum steady current, in amperes, that this wire can carry
without heating any part of the plastic above its operating limit of
200°F?
The thermal and
electrical conductivities may be assumed constant at the values given here:
Copper
220 5.1
x
lo5
Plastic
0.20 0.0
Answer:
13.4
amp
10A.5.
Free convection velocity.
(a) Verify the expression for the average velocity in the upward-moving stream in Eq.
10.9-16.
(b)
Evaluate
p
for the conditions given below.
(c)
What is the average velocity in the upward-moving stream in the system described in Fig.
10.9-1
for air flowing under these conditions?
Pressure
1
atm
Temperature of the heated wall
100°C
Temperature of the cooled wall
20°C
Spacing between the walls
0.6
cm
Answer:
2.3
cm/s
Problems
321
Plastic panel has
thermal conductivity
k
=
0.075
Btu/hr
ft
.
"F
(average value between
TI
and
T2)
Fig.
10A.6.
Determination of the thermal resistance of
a wall.
10A.6.
Insulating power of a wall (Fig. 10A.6). The "insulating power" of a wall can be measured
by means of the arrangement shown in the figure. One places a plastic panel against the wall.
In the pane1 two thermocouples are mounted flush with the panel surfaces. The thermal con-
ductivity and thickness of the plastic panel are known. From the measured steady-state tem-
peratures shown in the figure, calculate:
(a) The steady-state heat flux through the wall (and panel).
(b) The "thermal resistance" (wall thickness divided by thermal conductivity).
Answers:
(a)
14.3 Btu/hr ft2; (b)
4.2
f?.
hr
.
F/Btu
10A.7.
Viscous heating
in
a ball-point pen. You are asked to decide whether the apparent decrease
in viscosity in ball-point pen inks during writing results from "shear thinning" (decrease in
viscosity because of non-Newtonian effects) or "temperature thinning" (decrease in viscosity
because of temperature rise caused
by
viscous heating). If the temperature rise is less than lK,
then "temperature thinning" will not be important. Estimate the temperature rise using
Eq.
10.4-9 and the following estimated data:
Clearance between ball and holding cavity 5
X
in.
Diameter of ball
I
mrn
Viscosity of ink
lo4
CP
Speed of writing
100 in. /min
Thermal conductivity of ink (rough guess)
5
X
cal/s cm
.
C
10A.8.
Temperature rise in an electrical wire.
(a)
A copper wire, 5 mm in diameter and 15 ft long, has a voltage drop of 0.6 volts. Find the
maximum temperature in the wire if the ambient air temperature is
25°C
and the heat transfer
coefficient
h
is
5.7
Btu/hr.
f@.
F.
(b) Compare the temperature drops across the wire and the surrounding air.
10B.l.
Heat conduction from a sphere to a stagnant fluid.
A
heated sphere of radius
R
is sus-
pended
in
a
large, motionless body of fluid. It is desired to study the heat conduction in the
fluid surrounding the sphere in the absence of convection.
(a) Set up the differential equation describing the temperature Tin the surrounding fluid as a
function of
r,
the distance from the center of the sphere. The thermal conductivity
k
of the
fluid is considered constant.
(b) Integrate the differential equation and use these boundary conditions to determine the in-
tegration constants: at
r
=
R,
T
=
TR;
and at
r
=
w,
T
=
T,.
322
Chapter
10
Shell Energy Balances and Temperature Distributions in Solids and Laminar Flow
Fig.
10B.3.
Temperature distribution in a cylindrical fuel-
rod assembly.
(c)
From the temperature profile, obtain an expression for the heat flux at the surface.
Equate
this result to the heat flux given by "Newton's law of cooling" and show that a dimensionless
heat transfer coefficient (known as the
Nusselt number)
is given by
in which
D
is the sphere diameter. This well-known result provides the limiting value of
Nu
for heat transfer from spheres at low Reynolds and Grashof numbers (see
s14.4).
(d)
In what respect are the Biot number and the Nusselt number different?
10B.2.
Viscous heating in slit flow.
Find the temperature profile for the viscous heating problem
shown in Fig.
10.4-2,
when given the following boundary conditions: at
x
=
0,
T
=
T,;
at
x
=
b,
qx
=
0.
Answer:
-
-
10B.3
Heat conduction in a nuclear fuel rod assembly
(Fig.
10B.3).
Consider a long cylindrical
nu-
clear fuel rod, surrounded by an annular layer of aluminum cladding. Within the fuel
rod
heat is produced by fission; this heat source depends on position approximately as
Here
Sno
and
b
are known constants, and
r
is the radial coordinate measured from the axis
of
the cylindrical fuel rod. Calculate the maximum temperature in the fuel rod if the outer
sur-
face of the cladding is in contact with a liquid coolant at temperature
TL.
The heat transfer
co-
efficient at the cladding-coolant interface is
h,,
and the thermal conductivities of the fuel
rod
and cladding are
k,
and
kc.
Answer:
TF,max
-
TL
=
-
10B.4.
Heat conduction in
an
annulus
(Fig.
10B.4).
(a)
Heat is flowing through an annular wall of inside radius
r,,
and outside radius
r,.
The
thermal conductivity varies linearly with temperature from
ko
at
To
to
k,
at
TI.
Develop an
ex-
pression for the heat flow through the wall.
(b)
Show how the expression in (a) can be simplified when
(r,
-
r,)/r,
is very small. Interpret
the result physically.
Problems
323
'i!
Fig.
10B.4.
Temperature profile in an annular wall.
10B.5.
Viscous heat generation in a polymer melt. Rework the problem discussed in 510.4 for a
molten polymer, whose viscosity can be adequately described by the power law model (see
Chapter
8).
Show that the temperature distribution is the same as that in
Eq.
10.4-9 but with
the Brinkrnan number replaced by
Br,
=
10B.6.
Insulation thickness for
a
furnace wall (Fig. 10B.6).
A
furnace wall consists of three layers:
(i)
a layer of heat-resistant or refractory brick, (ii) a layer of insulating brick, and (iii) a steel
plate, 0.25 in. thick, for mechanical protection. Calculate the thickness of each layer of brick to
give minimum total wall thickness if the heat loss through the wall is to be 5000 ~tu/ft~
hr,
assuming that the layers are in excellent thermal contact. The following information is
available:
Maximum Thermal conductivity
allowable (Btu/hr
ft
-
F)
Material temperature at 100°F at 2000°F
Refractory brick 2600°F 1.8 3.6
Insulating brick 2000°F
0.9 1.8
Steel
-
26.1
-
Answer:
Refractory brick, 0.39 ft; insulating brick, 0.51 ft.
10B.7.
Forced-convection heat transfer in flow between parallel plates (Fig. 10B.7).
A
viscous fluid
with temperature-independent physical properties is in fully developed laminar flow
be-
tween two flat surfaces placed a distance
2B
apart. For
z
<
0 the fluid temperature is uniform
at
T
=
T,. For
z
>
0 heat is added at a constant, uniform flux
qo
at both walls. Find the temper-
ature distribution T(x,
z)
for large
z.
(a) Make a shell energy balance to obtain the differential equation for T(x,
z).
Then discard
the viscous dissipation term and the axial heat conduction term.
Steel
plate,
324
Chapter
10
Shell Energy Balances and Temperature Distributions in Solids and Laminar Flow
Fig.
10B.7.
Laminar, incompressible flow
between parallel plates, both of which are
being heated by a uniform wall heat flux
q,
starting at
z
=
0.
(b)
Recast the problem in terms of the dimensionless quantities
(c)
Obtain the asymptotic solution for large
z:
10B.8.
Electrical heating
of
a
pipe
(Fig.
10B.8).
In the manufacture of glass-coated steel pipes,
it
is
common practice first to heat the pipe to the melting range of glass and then to contact the
hot
pipe surface with glass granules. These granules melt and wet the pipe surface to form
a
tightly adhering nonporous coat. In one method of preheating the pipe, an electric current is
passed along the pipe, with the result that the pipe is heated (as in
g10.2).
For the purpose
of
this problem make the following assumptions:
(i)
The electrical conductivity of the pipe
k,
is constant over the temperature range
of
in-
terest. The local rate of electrical heat production
S,
is then uniform throughout the pipe wall.
(ii)
The top and bottom of the pipe are capped in such
a
way that heat losses through
them are negligible.
(iii)
Heat loss from the outer surface of the pipe to the surroundings is given by New-
ton's law of cooling:
q,
=
h(T,
-
T,).
Here
h
is a suitable heat transfer coefficient.
How much electrical power is needed to maintain the inner pipe surface at some desired
temperature, TK, for known
k,
Tat
h,
and pipe dimensions?
1-
I
Ambient air
L
1-
temperature
Ta
1
Pipe
wall
Electrical heating of a pipe.
-
Fig.
10B.8.
Problems
325
rR2(1
-
K~)L(T~
-
To)
Answer:
P
=
10B.9.
Plug
flow with forced-convection heat transfer. Very thick slurries and pastes sometimes
move in channels almost as a solid plug. Thus, one can approximate the velocity by a con-
stant value
v,
over the conduit cross section.
(a) Rework the problem of 510.8 for plug flow in a circular tube of radius
R.
Show that the
temperature distribution analogous to Eq.
10.8-31
is
in which
[
=
~Z/~?~V,R~,
and
@
and are defined as
in
510.8.
(b) Show that for plug flow in a plane slit of width 2B the temperature distribution analogous
to Eq. 10B.7-4 is
in which
5
=
k~/~~,v,~~, and
O
and
a
are defined as in Problem 108.7.
10B.lO.
Free convection in an annulus of finite height (Fig. 10B.10).
A
fluid is contained in a vertical
annulus closed at the top and bottom. The inner wall
of
radius KR
is
maintained at the tem-
perature
T,,
and the outer wall of radius R is kept at temperature
TI.
Using the assumptions
and approach of 510.9, obtain the velocity distribution produced by free convection.
(a) First derive the temperature distribution
in
which
l=
r/R.
(b) Then show that the equation of motion is
in which
A
=
(~~/p)(dp/dz
+
pig)
and
B
=
((p1gp,AT)R2/p In
K)
where AT
=
TI
-
T,.
(c)
Integrate the equation of motion (see
Eq.
C.l-11) and apply the boundary conditions to
evaluate the constants of integration. Then show that
A
can be evaluated
by
the requirement
of no net mass flow through any plane
z
=
constant, with the final resdt that
Fig.
10B.10.
Free convection pattern in an annular space
with
TI
>
T,.
326
Chapter 10
Shell Energy Balances and Temperature Distributions in Solids and Laminar Flow
Free convection with temperature-dependent viscosity. Rework the problem in 510.9, tak-
ing into account the variation of viscosity with temperature. Assume that the "fluidity" (reci-
procal of viscosity) is the following linear function of the temperature
Use the
q,
&,
and Gr defined in 510.9 (but with
ji
instead of
p)
and in addition
b,
=
:PAT,
b,
=
%,AT
and
P
=
and show that the differential equation for the velocity distribution is
Follow the procedure in 510.9, discarding terms containing the third and higher powers
of
AT. Show that this leads to
P
=
&
Grb,
+
&
Grb, and finally:
Sketch the result to show how the velocity profile becomes skewed because of the tempera-
ture-dependent viscosity.
Heat conduction with temperature-dependent thermal conductivity (Fig. 108.12).
The
curved surfaces and the end surfaces (both shaded in the figure) of the solid in the shape
of
a
half-cylindrical shell are insulated. The surface
0
=
0,
of area (r2
-
r,)L, is maintained at tem-
perature To, and the surface at
8
=
T,
also of area (r,
-
rJL, is kept at temperature
T,.
The thermal conductivity of the solid varies linearly with temperature from
ko
at T
=
To
tok,at
T=
T,.
(a) Find the steady-state temperature distribution.
(b)
Find the total heat flow through the surface at
8
=
0.
Flow reactor with exponentially temperature-dependent source. Formulate the function
F(O)
of
Eq.
10.5-7 for a zero-order reaction with the temperature dependence
in which
K
and
E
are constants, and
R
is the gas constant. Then insert
F(O)
into Eqs. 10.5-15
through 20 and solve for the dimensionless temperature profile with
kz,e,
neglected.
Evaporation loss from an oxygen tank.
(a) Liquefied gases are sometimes stored in well-insulated spherical containers vented to
the
atmosphere. Develop an expression for the steady-state heat transfer rate through the walls
of
such a container, with the radii of the inner and outer walls being r, and
r,
respectively and
~urface'L
r2
at
T,
1-r
'\
Surface
at
To
Fig.
10B.12.
Tangential heat conduction in an annular shell.